Question

Inverse trigonometry question doubt.
\LARGE f(x)=\cos^{-1}x+\cos^{-1}(\frac{x}{2}+\1/2 sqrt{3-3x^2})

Answer 1

\[\LARGE f(x)=\cos^{-1}x+\cos^{-1}(\frac{x}{2}+\sqrt{3-3x^2})\]
Then
A)f(2/3)=pi/3
B)f(2/3)= 2 arccos(2/3)-pi/3
C)f(1/3)=pi/2
D)f(1/3)=2 arccos(1/3)-pi/3

Answer 2

I tried to solve the 2nd part by substituting x=sin theta
It gave me,
\[\cos^{-1}(\frac{1}{2}\sin \theta+\frac{\sqrt{3}}{2}\cos \theta)=\cos^{-1}(\sin(\frac{\pi}{3}+\theta))=\frac{\pi}{2}-\sin^{-1}(\sin(\frac{\pi}{3}+\theta)\]

Answer 3

sorry thats should be (pi/6+theta)

Answer 4

\[=>\frac{\pi}{2}-\frac{\pi}{6}-\theta=\frac{\pi}{3}-\sin^{-1}x\]

Answer 5

why do you think it's \(\pi/6\)?

Answer 6

hmm its not :P

Answer 7

its sin ( pi/3+theta) only

Answer 8

i got confused with smt else ,nvm.

Answer 9

\[\LARGE \frac{\pi}{2}-\frac{\pi}{3}-\theta=\frac{\pi}{6}-\sin^{-1}x\]

Answer 10

\[\Large =>\cos^{-1}x-\sin^{-1}+\frac{\pi}{6}=\cos^{-1}-(\frac{\pi}{2}-\cos^{-1}x+\frac{\pi}{6})\]

Answer 11

cos^-1x*

Answer 12

\[\LARGE 2\cos^{-1}x- \frac{2 \pi}{3}\]
is this much correct?

Answer 13

@oldrin.bataku

Answer 14

can anybody hear me...or am I talking to myself..

Answer 15

my mind is running empty..in this search for someone else..

Answer 16

who doesn't look right through me..its all just static in my head..

Answer 17

can anybody tell me..why..im lonely like a satellite..

Answer 18

did you mean$$\LARGE f(x)=\cos^{-1}x+\cos^{-1}(\frac{x}{2}+\frac12\sqrt{3-3x^2})$$

Answer 19

that is what i wrote?

Answer 20

no this has a \(1/2\) before the radical

Answer 21

oh sorry 1/2 is missing sorry,1/2 is there too

Answer 22

Substituting \(x=\cos t\):$$\begin{align*}f(\cos t)&=t+\arccos\left(\frac12\cos t+\frac{\sqrt3}2\sin t\right)\\&=t+\arccos(\sin(\pi/6+t))\\&=t+\arccos(\cos(\pi/2+\pi/6+t))\\&=2t+2\pi/3\end{align*}$$for \(t\in[0,\pi/3)\) hence for \(x\in(1/2,1]\) we have:$$f(x)=2\arccos(x)+2\pi/3\\f(2/3)=2\arccos(2/3)+2\pi/3$$ which is neither option

Answer 23

ok so now we continue

Answer 24

so i did correct?

Answer 25

oh a difference of sign :O

Answer 26

using f(2/3) I get one of your options.